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The question appears to be about finding a point that is equidistant from all vertices of a pyramid \( S.ABCD \), where the base \( ABCD \) is a rhombus, the edge \( SA \) is perpendicular to the base, and \( BD = 2AC \).

To break it down:

1. **The Shape and Geometry**: The pyramid has a rhombus as its base, and the edge \( SA \) is perpendicular to the base. Additionally, one diagonal of the rhombus, \( BD \), is twice as long as the other diagonal, \( AC \). This means the base is symmetric along the line connecting the center of the rhombus \( I \) and the apex \( S \).

2. **The Concept of an Equidistant Point**: The question asks for the point that is equidistant from all vertices of the pyramid. Such a point would be the center of a sphere that circumscribes the pyramid, called the "circumsphere". The center of the circumsphere would need to be equidistant from the vertices \( S, A, B, C, D \).

3. **Options**:
   - **A. Trung điểm SC**: The midpoint of \( SC \).
   - **B. Không tồn tại điểm cách đều các đỉnh của hình chóp**: This option suggests no such point exists.
   - **C. Điểm nằm trên đường thẳng d // SA**: A point on a line parallel to \( SA \).
   - **D. Trung điểm SD**: The midpoint of \( SD \).

4. **Analysis**:
   - Since the base is a rhombus and \( SA \) is perpendicular to the base, the center of the base \( I \), which is the intersection of the diagonals \( BD \) and \( AC \), seems like a candidate for the equidistant point from the base vertices \( A, B, C, D \). 
   - The height \( SA \) is crucial because it suggests symmetry along the vertical axis. However, finding a point equidistant from all vertices, including the apex \( S \), may require a more sophisticated geometric property.
   - It is possible that no such point exists, especially if the geometry does not allow for a circumcenter in 3D space that satisfies the condition for all five vertices.

Thus, the most plausible answer could be **B. không tồn tại điểm cách đều các đỉnh của hình chóp**, meaning no such equidistant point exists.

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Would you like more details on why no equidistant point exists, or have any further questions? Here are some related questions for deeper understanding:

1. What is a circumsphere in a 3D geometric solid?
2. How do the properties of a rhombus affect a pyramid's geometry?
3. Can every pyramid have a point equidistant from all its vertices?
4. What is the role of symmetry in determining equidistant points?
5. How do diagonals of a rhombus affect its 3D structure?

**Tip**: In problems involving pyramids, always analyze the symmetry of both the base and the apex to better understand the geometry.

Cho hình chóp S.ABCD có đáy ABCD là hình thoi tâm I, cạnh bên SA vuông góc với đáy, BD = 2AC. Điểm cách đều các đỉnh của hình chóp là?

This geometry problem involves a pyramid S.ABCD with a rhombus base, where edge SA is perpendicular to the base and BD = 2AC. The challenge is to determine whether a point exists that is equidistant from all vertices of the pyramid.

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